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Which one satisfies the equation $\sqrt[3]{6+\sqrt x} + \sqrt[3]{6-\sqrt x} = \sqrt[3] {3}$

(A)$27$ (B)$32$ (C)$45$ (D)$52$ (E)$63$

let $a = 6+\sqrt x , b=6-\sqrt x$

cube each side

\begin{align} (\sqrt[3]a + \sqrt[3]b)^3 &= (\sqrt[3]3)^3 \\ (\sqrt[3]{a^2} + 2\sqrt[3]{ab} + \sqrt[3]{b^2})(\sqrt[3]a + \sqrt[3]b) &= 3 \\ \sqrt[3]{a^3} + \sqrt[3]{3a^2b} + \sqrt[3]{3ab^2} + \sqrt[3]{b^3} &= 3 \\ a + b + 3\sqrt[3]{a^2b} + 3\sqrt[3]{ab^2} &= 3 \end{align}

There's still had cube root, how do I remove it?

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  • $\begingroup$ Note that $$\displaystyle (\sqrt[3]{a} + \sqrt[3]{b} )^3 = (\sqrt[3]{a} + \sqrt[3]{b}) ( a^{\frac{2}{3}} + 2 a^{\frac{1}{3}} b^{\frac{1}{3}} + b^{\frac{2}{3}})$$ not $(\sqrt[3]{a^2} + \sqrt[3]{2ab} + \sqrt[3]{b^2})(\sqrt[3]a + \sqrt[3]b)$, and $ab = 36 - x$. $\endgroup$
    – IrbidMath
    Jul 28 '15 at 13:11
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\begin{align*} a + b + 3\sqrt[3]{a^2b} + 3\sqrt[3]{ab^2} &= 3 \\ 6+\sqrt{x} + 6-\sqrt{x} + 3\sqrt[3]{a^2b} + 3\sqrt[3]{ab^2} &= 3 \\ 12 + 3\sqrt[3]{a^2b} + 3\sqrt[3]{ab^2} &= 3 \\ 3\sqrt[3]{a^2b} + 3\sqrt[3]{ab^2} &= -9 \end{align*}

Divide by $3\sqrt[3]{ab}$:

$$\sqrt[3]{b} + \sqrt[3]{a} = \frac{-9}{3\sqrt[3]{ab}}$$

Using the original equation we see that the right handed side equals $\sqrt[3]{3}$. We use that. Then substituting $a=6+\sqrt{x}$, $b=6-\sqrt{x}$ will yield a simple lineair equation in $x$ which we can solve.

\begin{align*} \frac{-9}{3\sqrt[3]{ab}} &= \sqrt[3]{3} \\ \frac{-729}{27ab}&=3 \\ ab &=-9 \\ \\ (6+\sqrt{x})(6-\sqrt{x}) &=-9 \\ 36-x &= -9 \\ x &= 45 \end{align*}

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Step 1 (conjecture): there is some $e>0$ such that $$ 6+\sqrt{x}=(e+\sqrt[3]{3}/2)^3,\quad 6-\sqrt{x}=(-e+\sqrt[3]{3}/2)^3.\tag{$*$} $$ You can see that if such an $e$ exists then the equality $\sqrt[3]{6+\sqrt x} + \sqrt[3]{6-\sqrt x} = \sqrt[3] {3}$ is satisfied. Solving for $e$ is simple: $$ 12=(6+\sqrt{x})+(6-\sqrt{x})=(e+\sqrt[3]{3}/2)^3+(-e+\sqrt[3]{3}/2)^3=\frac{3}{4}+3\sqrt[3]{3}e^2 $$ from which $e=\frac{1}{2}\sqrt[3]{3}\sqrt{5}$.

Step 2 (verify): with $e$ as above, you can verify that ($*$) is satisfied with $\sqrt{x}=3\sqrt{5}$ which is equivalent to $x=45$.

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    $\begingroup$ An elegant approach!(+1) $\endgroup$ Jul 28 '15 at 13:26
  • $\begingroup$ @user3313320 Thanks. I was trying to reverse engineer the problem. :) $\endgroup$ Jul 28 '15 at 13:27
  • $\begingroup$ The problem is quite symmetric :) $\endgroup$ Jul 28 '15 at 13:28
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Let $ a = \sqrt[3]{6+\sqrt x} $ and $ b = \sqrt[3]{6-\sqrt x}$, then we have:

$$ \left\{\begin{matrix} a^3 + b^3 = 12 \\ a + b = \sqrt[3]{3} \end{matrix}\right. $$ Therefore,

$$ ab = \sqrt[3]{-9}$$

or, $$ \sqrt[3]{36 - x} = \sqrt[3]{-9} $$ then,

$$ x = 45$$

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  • $\begingroup$ few years later after looking back, i found this one is the easiest answer $\endgroup$
    – wuiyang
    Nov 30 '17 at 9:36
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Start with your idea $a=6+\sqrt{x}$ and $b=6-\sqrt{x}$. You have $$ab=36-x \text{, } \sqrt[3]{a} + \sqrt[3]{b} = \sqrt[3]{3} \text{ and } a+b=12$$ Raise the middle equation to power $3$ you get $$a+b +3\sqrt[3]{ab}(\sqrt[3]{a} + \sqrt[3]{b})=3$$ and using the initial assumption $$12 +3\sqrt[3]{ab}\sqrt[3]{3}=3 \text{ or } \sqrt[3]{ab}\sqrt[3]{3}=-3$$ and finally $$ab=-9$$ which allows to conclude to $x=45$ as $ab=36-x$.

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go to power of three both side $$(\sqrt[3]{6+\sqrt x} + \sqrt[3]{6-\sqrt x} = \sqrt[3] {3})^3 \\\xrightarrow[(a+b)^3=a^3+b^3+3ab(a+b)]{} 6+\sqrt{x} +6-\sqrt{x} +3(\sqrt[3]{6+\sqrt x})(\sqrt[3]{6-\sqrt x})(\sqrt[3]{6+\sqrt x} + \sqrt[3]{6-\sqrt x} \sqrt[3] {3})=3\\ $$ we can subsitute $\sqrt[3]{6+\sqrt x} + \sqrt[3]{6-\sqrt x} = \sqrt[3] {3}$ so we have $$ 6+\sqrt{x} +6-\sqrt{x} +3(\sqrt[3]{6+\sqrt x})(\sqrt[3]{6-\sqrt x})\sqrt[3] {3}=3\\$$simplifying $$ 3(\sqrt[3]{6+\sqrt x})(\sqrt[3]{6-\sqrt x})\sqrt[3] {3}=3-12\\(\sqrt[3]{36- x})\sqrt[3] {3}=-3$$ to the power of three $$ 3(36- x)=-27\\36-x=-9\\x=45$$

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$$6-\sqrt{x} = t^3$$ then $$6+\sqrt{x} = t^3+2\sqrt{x}$$ The expression becomes: $$t + \sqrt[3]{t^3+2\sqrt{x}} = \sqrt[3]{3}$$ Substitute back $\sqrt{x}$ in terms of t, rewrite so the cube root is alone. Raise to 3, solve 3rd degree polynomial equation. Test all roots.


Fun fact: The valid $t$ solution happens to become closely related to the golden ratio: $t =\sqrt[3]{3}\left(\frac{1+\sqrt{5}}{2}\right) = \sqrt[3]{3}\varphi$, where $\varphi = \frac{\sqrt{5}+1}{2}$

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